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Because this puzzle has just been asked, how about a similar puzzle.

Change TWO pixels to make this equation true.

(10 + 1) * (10 - 1) = 101

source

NOTE:

I didn't intend on there being so many possible solutions, you guys are pretty clever! I'll accept the answer I sought but still upvote any additional solution.

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5 Answers 5

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Change the equation to

$$(10+i)(10-i)=101$$ where $i$ is the imaginary number such that $i^2=-1$.

It looks like this:
Image of solution

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This puzzle was fun.

Turning on two pixels, in binary we get:

Hint: 10.1

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    $\begingroup$ $+1$ Clever! However, there are more solutions! $\endgroup$ Commented Mar 30 at 5:32
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    $\begingroup$ I'm not seeing it, does it say (10 + 0.1)*(10 - 1)? Because for me that returns 90.9? $\endgroup$ Commented Mar 31 at 7:18
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    $\begingroup$ @Excellor Both the left and right side of the equation are written in base 2. $\endgroup$ Commented Mar 31 at 8:48
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    $\begingroup$ @Will.Octagon.Gibson Oh, I missed "in binary we get"; but thanks for the explanation! Clear now. $\endgroup$ Commented Mar 31 at 9:27
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Oh! I can do it in 1 pixel!

(10+1)-(10-.1)=|0| Just note that the right hand side doesn't read 101 but |0|. An easy thing to miss given the unfortunate font choice on this display. :)

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    $\begingroup$ Toggling the mulitiply pixel would match the 2 pixels requirement. $\endgroup$ Commented Mar 30 at 17:38
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    $\begingroup$ @z100 Or I can do the same trick in the first parenthesis as well. :) TBH it was pretty clear that this was not the intended solution, so I posted this more as a joke. :) $\endgroup$ Commented Mar 30 at 21:56
  • $\begingroup$ Rot13(Nethnoyl lbh pbhyq nyfb qb vg va mreb cvkryf, nffhzvat lbh pbafvqre 0+ naq 0- nf ahzoref. 0+ naq 0- ner hfrq jura jbexvat jvgu yvzvgf.) $\endgroup$ Commented Mar 31 at 20:11
  • $\begingroup$ @Brian - I suppose, but that would be odd in this context. $\endgroup$ Commented Mar 31 at 22:19
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With a slight misrendering because I begin a number with $0$:

$(10+1)(1.0-1)=!01$

where $!$ before a number indicates number of derangements (which do not exist with the number of objects specified on the right side above).

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  • $\begingroup$ I'm hesitant to agree as DNE is not the same as 0... $\endgroup$ Commented Mar 31 at 23:07
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    $\begingroup$ I am saying the derangements do not exist with one object, in which case their number is zero. $\endgroup$ Commented Mar 31 at 23:12
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I can also do it in 1 pixel:

$(10+1)\cdot(!0-1)=|0|$
The $!0$ might mean the derangement (which equals 1) or the boolean not of 0 (which is true so it equals 1)

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